It is well-known that the Hausdorff dimension is invariant under bi-Lipschitz mappings. I would be interested in a specific converse of this invariance. Let $X$ and $Y$ be two CAT(0) spaces having the same Hausdorff dimension.
In addition, we assume that the metric curvature of $X$ and $Y$ is bounded from below. More precisely, we assume that we have a $\kappa<0$ such that the metric curvature is at least $\kappa$. In order to explain this, we need some preparation. We denote by $M_\kappa$ the hyperbolic plane where the metric is scaled by $1/\sqrt{-\kappa}$. For a triple of points $x,y,z \in X$ we may pick points $x',y',z'\in M_\kappa$ where $d_X(x,y)=d_{M_\kappa}(x,y)$, $d_X(x,z) = d_{M_\kappa}(x,z)$ and $d_X(y,z)=d_{M_\kappa}(y,z)$. For two points $a,b$ we denote by $[a,b]$ a geodesic connecting $a$ and $b$. For a point $u\in [x,y]$ we call a point $u'\in [x',y']$ a comparison point if $d_X(x,u)=d_{M_\kappa}(x',u')$ and $d_X(u,y)=d_{M_\kappa}(u,y)$. Now the condition we have is that for every triple $x,y,z\in X$ and all $u\in[x,y]$ and $v\in [x,z]$, we have $$ d_X(u,v) \geq d_{M_\kappa}(u',v') $$ where $u'\in[x',y']$ and $v'\in [x',z']$ are comparison points for $u$ and $v$, respectively.
Can we find two open balls $B_1 \subset X$ and $B_2\subset Y$ and a bi-Lipschitz surjective mapping $f\colon B_1\to B_2$?
Without the lower bound on the curvature, this is not possible as pointed out by YCor and user142382. One reason is that there are compact $\mathbb{R}$-trees of arbitrary large Hausdorff dimension - see e.g.
P. D. Andreev and V. N. Berestovskiĭ: Dimensions of ℝ-trees and self-similar fractal spaces of nonpositive curvature. Siberian Advances in Mathematics 17: 79–90(2007). DOI: 10.3103/S1055134407020010
